A321516 -id:A321516 - cp's OEIS frontend

A321517 Numbers k such that A294902(k) != A321516(k).

50, 75, 100, 125, 150, 175, 190, 200, 222, 225, 238, 242, 246, 250, 275, 285, 300, 325, 333, 338, 350, 357, 363, 369, 374, 375, 380, 400, 425, 438, 442, 444, 450, 475, 476, 484, 492, 494, 500, 507, 525, 550, 555, 561, 570, 575, 578, 595, 600, 605, 615, 625

Offset: 1

Felix Fröhlich, Nov 12 2018

nonn

Cf. A033942, A294902, A321516.

A321516[n_] := Length[Select[Most[Divisors[n]], CompositeQ]]; abQ[n_] := DivisorSum[n, Total[IntegerDigits[#, 2]]*(-1)^Boole[#==n]&]>0; A294902[n_] := Length[Select[Most[Divisors[n]], abQ[#] &]]; Select[Range[650], A321516[#] != A294902 [#] &] (* Amiram Eldar, Nov 12 2018 after Jean-François Alcover at A175526 *)

a292257(n) = sumdiv(n, d, (dAntti Karttunen in A292257
a294905(n) = (a292257(n) <= hammingweight(n)) \\ after Antti Karttunen in A294905
a294902(n) = sumdiv(n, d, (dAntti Karttunen in A294902
a321516(n) = my(d=divisors(n), i=0); for(k=2, #d-1, if(!ispseudoprime(d[k]), i++)); i
is(n) = a294902(n)!=a321516(n)

A363159 a(1)=1. Thereafter, if a(n-1) is a novel term, a(n) is the smallest prime which does not divide a(n-1). If a(n-1) has been seen k (>1) times already then a(n) = k*a(n-1).

Original entry on oeis.org

1, 2, 3, 2, 4, 3, 6, 5, 2, 6, 12, 5, 10, 3, 9, 2, 8, 3, 12, 24, 5, 15, 2, 10, 20, 3, 15, 30, 7, 2, 12, 36, 5, 20, 40, 3, 18, 5, 25, 2, 14, 3, 21, 2, 16, 3, 24, 48, 5, 30, 60, 7, 14, 28, 3, 27, 2, 18, 36, 72, 5, 35, 2, 20, 60, 120, 7, 21, 42, 5, 40, 80, 3, 30, 90, 7, 28, 56, 3, 33, 2, 22, 3, 36, 108

Offset: 1

David James Sycamore, Jul 08 2023

nonn

It follows from the definition that the sequence is infinite. Every number appears multiple times according to its prime factorization. All primes p appear infinitely many times, prime powers p^k (k>1) appear once only, all squarefree semiprimes appear twice, and so on.

On the first occasion of A007947(a(n-1)) = A002110(k-1), a(n) is the first occasion of prime(k).

			a(2)=2 since 1 is a novel term and 2 is the least prime which does not divide 1, a(3)=3 since 3 is the smallest prime which does not divide 2.
a(4)=4 since 2 has appeared twice.
a(7) = 6, therefore a(8) = 5.
f(30) = A001221(30) + 1 since f(15)=2 and 2*15=30. No other divisor d of 30 has the property d*f(d) >= 30 thus f(30)=3+1=4.

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^20.

Cf. A001221, A002110, A007947, A053669, A321516, A351495, A359804.

nn = 120; c[_] := 0; a[1] = c[1] = k = 1;
Do[If[c[j] == 0,
     c[j]++; p = 2; While[Divisible[j, p], p = NextPrime[p]]; Set[k, p],
     c[j]++; Set[k, j c[j]] ];
  Set[{a[n], j}, {k, k}], {n, 2, nn}];
Array[a, nn] (* Michael De Vlieger, Jul 08 2023 *)

lista(nn) = my(c, p, v=vector(nn)); v[1]=1; for(k=2, nn, if(c=sum(i=1, k-2, v[i]==v[k-1]), v[k]=(c+1)*v[k-1], p=2; while(v[k-1]%p==0, p=nextprime(p+1)); v[k]=p)); v \\ Jinyuan Wang, Jul 11 2023

For integer m let f(m) be the number of times m appears in the sequence.
f(1)=1, f(p)->oo for all prime p, and for n composite the following recursion applies:
f(n) = A001221(n) + Sum_{i=1..A321516(n)} [k_i*f(k_i)>=n], where k_i is a composite divisor of n and [] is the Iverson bracket.

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A321517 Numbers k such that A294902(k) != A321516(k).

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A363159 a(1)=1. Thereafter, if a(n-1) is a novel term, a(n) is the smallest prime which does not divide a(n-1). If a(n-1) has been seen k (>1) times already then a(n) = k*a(n-1).

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